an object projected vertically upwards with initial velocity 40 m/s from the ground (g=10m/s) time of the flight is

To find the time of flight of an object projected vertically upwards, we can use the equation of motion:

s = ut - (1/2)gt^2

Where: s = displacement (in this case, the height reached by the object) u = initial velocity g = acceleration due to gravity t = time of flight

In this case, the initial velocity (u) is 40 m/s and the acceleration due to gravity (g) is 10 m/s^2.

Since the object is projected vertically upwards, the final displacement (s) will be zero when it reaches its maximum height and starts falling back down.

Therefore, we can rewrite the equation as:

0 = (40)t - (1/2)(10)t^2

Simplifying the equation, we get:

0 = 40t - 5t^2

Rearranging the equation, we have:

5t^2 - 40t = 0

Factoring out a common factor of t, we get:

t(5t - 40) = 0

Setting each factor equal to zero, we have two possible solutions:

t = 0 (which is not possible in this case) 5t - 40 = 0

Solving for t, we find:

5t = 40 t = 40/5 t = 8 seconds

Therefore, the time of flight for the object projected vertically upwards with an initial velocity of 40 m/s is 8 seconds.